## Tag `02NW`

Chapter 28: Morphisms of Schemes > Section 28.48: Generically finite morphisms

Lemma 28.48.1. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $\eta \in Y$ be a generic point of an irreducible component of $Y$. The following are equivalent:

- the set $f^{-1}(\{\eta\})$ is finite,
- there exist affine opens $U_i \subset X$, $i = 1, \ldots, n$ and $V \subset Y$ with $f(U_i) \subset V$, $\eta \in V$ and $f^{-1}(\{\eta\}) \subset \bigcup U_i$ such that each $f|_{U_i} : U_i \to V$ is finite.
If $f$ is quasi-separated, then these are also equivalent to

- (3) there exist affine opens $V \subset Y$, and $U \subset X$ with $f(U) \subset V$, $\eta \in V$ and $f^{-1}(\{\eta\}) \subset U$ such that $f|_U : U \to V$ is finite.
If $f$ is quasi-compact and quasi-separated, then these are also equivalent to

- (4) there exists an affine open $V \subset Y$, $\eta \in V$ such that $f^{-1}(V) \to V$ is finite.

Proof.The question is local on the base. Hence we may replace $Y$ by an affine neighbourhood of $\eta$, and we may and do assume throughout the proof below that $Y$ is affine, say $Y = \mathop{\rm Spec}(R)$.It is clear that (2) implies (1). Assume that $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ is finite. Choose affine opens $U_i \subset X$ with $\xi_i \in U_i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ each of the morphisms $U_i \to Y$ is finite. In other words (2) holds.

It is clear that (3) implies (1). Assume $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ and assume that $f$ is quasi-separated. Since $Y$ is affine this implies that $X$ is quasi-separated. Since each $\xi_i$ maps to a generic point of an irreducible component of $Y$, we see that each $\xi_i$ is a generic point of an irreducible component of $X$. By Properties, Lemma 27.29.1 we can find an affine open $U \subset X$ containing each $\xi_i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ the morphisms $U \to Y$ is finite. In other words (3) holds.

It is clear that (4) implies all of (1) – (3) with no further assumptions on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to show that the equivalent conditions (1) – (3) imply (4). Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact. Hence $Z = X \setminus U$ is quasi-compact, hence the morphism $f|_Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that $\eta \not \in f(Z)$. Hence by Lemma 28.8.4 we see that there exists an affine open neighbourhood $V'$ of $\eta$ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset$. Then we have $f^{-1}(V') \subset U$ and this means that $f^{-1}(V') \to V'$ is finite. $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 11819–11845 (see updates for more information).

```
\begin{lemma}
\label{lemma-generically-finite}
Let $X$, $Y$ be schemes.
Let $f : X \to Y$ be locally of finite type.
Let $\eta \in Y$ be a generic point of an irreducible component
of $Y$. The following are equivalent:
\begin{enumerate}
\item the set $f^{-1}(\{\eta\})$ is finite,
\item there exist affine opens $U_i \subset X$, $i = 1, \ldots, n$
and $V \subset Y$ with $f(U_i) \subset V$,
$\eta \in V$ and $f^{-1}(\{\eta\}) \subset \bigcup U_i$
such that each $f|_{U_i} : U_i \to V$ is finite.
\end{enumerate}
If $f$ is quasi-separated, then these are also equivalent to
\begin{enumerate}
\item[(3)] there exist affine opens $V \subset Y$,
and $U \subset X$ with $f(U) \subset V$,
$\eta \in V$ and $f^{-1}(\{\eta\}) \subset U$
such that $f|_U : U \to V$ is finite.
\end{enumerate}
If $f$ is quasi-compact and quasi-separated,
then these are also equivalent to
\begin{enumerate}
\item[(4)] there exists an affine open $V \subset Y$, $\eta \in V$
such that $f^{-1}(V) \to V$ is finite.
\end{enumerate}
\end{lemma}
\begin{proof}
The question is local on the base. Hence we may replace $Y$ by an
affine neighbourhood of $\eta$, and we may and do assume throughout
the proof below that $Y$ is affine, say $Y = \Spec(R)$.
\medskip\noindent
It is clear that (2) implies (1).
Assume that $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ is finite.
Choose affine opens $U_i \subset X$ with $\xi_i \in U_i$.
By Algebra, Lemma \ref{algebra-lemma-generically-finite} we see
that after replacing $Y$ by a standard open in
$Y$ each of the morphisms $U_i \to Y$ is finite.
In other words (2) holds.
\medskip\noindent
It is clear that (3) implies (1).
Assume $f^{-1}(\{\eta\}) = \{\xi_1, \ldots, \xi_n\}$ and assume
that $f$ is quasi-separated.
Since $Y$ is affine this implies that $X$ is quasi-separated.
Since each $\xi_i$ maps to a generic point of an irreducible component
of $Y$, we see that each $\xi_i$ is a generic point of an irreducible
component of $X$.
By Properties, Lemma \ref{properties-lemma-maximal-points-affine}
we can find an affine open $U \subset X$ containing each $\xi_i$.
By Algebra, Lemma \ref{algebra-lemma-generically-finite} we see
that after replacing $Y$ by a standard open in
$Y$ the morphisms $U \to Y$ is finite.
In other words (3) holds.
\medskip\noindent
It is clear that (4) implies all of (1) -- (3) with no further assumptions
on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to
show that the equivalent conditions (1) -- (3) imply (4).
Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact
and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact.
Hence $Z = X \setminus U$ is quasi-compact, hence the morphism
$f|_Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that
$\eta \not \in f(Z)$. Hence by
Lemma \ref{lemma-quasi-compact-generic-point-not-in-image}
we see that there exists an affine open
neighbourhood $V'$ of $\eta$ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset$.
Then we have $f^{-1}(V') \subset U$ and this means
that $f^{-1}(V') \to V'$ is finite.
\end{proof}
```

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